Pls pls pls pls
EB−→− bisects ∠RED. Find the measure of ∠BED if ∠REB = 5x – 3 and ∠RED = 8x + 16

answer set (b) 0.5, 3/4 80% is in order from least to greatest.  

step-by-step explanation:

in order to compare rational numbers to each other, it is much easier if they are in the same form.   typically, the best form in which to evaluate rational numbers is in decimals.   in order to convert a percent to a decimal, you would divide the given percentage by 100.   you can also accomplish the conversion by moving the decimal point over two places to the left.   for example, the decimal value of 80% is 80/100, or 0.8.   in order to convert a fraction to a decimal, you need to divide the numerator by the denominator.   for example, 3/4 is 0.75.   so, for choice (b) 0.5 3/4 (0.75), 80% (0.8) is the only choice in which the numbers are in order from least to greatest.  

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.

56 44 2 4

step-by-step explanation: